Chapter 24, Problem 7P

To determine

The wavelength of visible light would have a minimum at the same location.

Answer to Problem 7P

  $613nm$

Explanation of Solution

Given info:

Wavelength, $\lambda =460nm$

Order, $m=2.$

Formula used:

It is known that $\sin \theta =\frac{m\lambda }{d}$

Where,

  $m$ is an integer

  $\lambda$ is wavelength

  $d$ is a distance

Calculation:

The following conditions must be met:

  $d\sin \left(\theta \right)=2\left(460\right)=920nm$

  $\lambda \left(m+0.5\right)=920$

Where, $m=0,1,2,3,\mathrm{...}$

Now, solve for $\lambda$ using various $m\text{'}s$

  $\lambda =\frac{920}{m+0.5}$

By putting the value $m=0\to \lambda =1840nm$

By putting the value $m=1\to \lambda =613nm$

By putting the value $m=2\to \lambda =368nm$

Conclusion: The required wavelength that is visible: $613nm$ .